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\title{Towards SDN IP FRR}
\author{yinqingwang }
\date{July 2014}

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\begin{document}

\maketitle
\tableofcontents
\begin{abstract}
This is a test abstract!

\end{abstract}
\section{Introduction}
\begin{CJK}{UTF8}{gkai}
   	
   	这是一个楷体中文测试，处理简体字。 Good time!
   	
\end{CJK}
\begin{CJK}{UTF8}{gbsn}
	
	这是一个宋体中文测试，处理简体字。 Good day!
	
\end{CJK}

\begin{CJK}{UTF8}{bkai}
	
	這是一個big5編碼的楷體中文測試，處理繁體文字。
	
\end{CJK}

\begin{CJK}{UTF8}{bsmi}
	
	這是一個个big5編碼的明體中文測試，處理繁體文字。
	
\end{CJK}


There is another theory which states that this has already happened.\citep{monsanto2013composing}

\section{Method-Inequality}

\textbf{Prove:} $\sqrt{a^2+bc} + \sqrt{b^2+ac} + \sqrt{c^2+ab}\leq{3\over2}(a+b+c)$, with a,b,c are nonnegative.
\\\textbf{Proof.} 

(i) if abc=0, W.L.O.G assume c=0, $\Longleftrightarrow a+b \geq 2\sqrt{ab}$, it is clearly true.

(ii) in the following, assume $abc > 0$.

\textbf{Case 1: $a^2+b^2+c^2 \leq 2(ab+bc+ca)$}

according to Cauchy-Schwarz inequality, we have:
$$LHS \leq \sqrt{3(a^2+b^2+c^2+ab+bc+ca)}$$
$$\Longleftrightarrow \sqrt{3(a^2+b^2+c^2+ab+bc+ca)} \leq {3\over2}(a+b+c)$$
$$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ca) \leq {9\over4}(a+b+c)^2$$
$$\Longleftrightarrow a^2+b^2+c^2 \leq 2(ab+bc+ca)$$

it is clearly true.

\textbf{Case 2: $a^2+b^2+c^2 \geq 2(ab+bc+ca)$}

W.L.O.G assume \textbf{c=max\{a,b,c\}},
according to Cauchy-Schwarz inequality, $$\sqrt{a^2+bc} + \sqrt{b^2+ac} \leq \sqrt{2(a^2+b^2+ac+bc)}$$

and $$\sqrt{c^2+ab}=\sqrt{c^2+ab}-c+c=\frac{ab}{\sqrt{c^2+ab}+c} +c \leq \frac{ab}{2c} +c$$

$LHS \leq \sqrt{2(a^2+b^2+ac+bc)} + \frac{ab}{2c} +c \leq {3\over2}(a+b+c)$

$$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b+c) - c - \frac{ab}{2c}$$
$$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b) + {1\over2}c - \frac{ab}{2c}$$
$$\Longleftrightarrow 2(a^2+b^2+ac+bc) \leq ({3\over2}(a+b) + {1\over2}c - \frac{ab}{2c})^2$$
$$\Longleftrightarrow \frac{a^2b^2-6a^2bc-6ab^2c+a^2c^2+16abc^2+b^2c^2-2ac^3-2bc^3+c^4}{4c^2} \geq 0$$
$$\Longleftrightarrow \frac{a^2b^2+abc(16c-6a-6b)+c^2(a^2+b^2+c^2-2ac-2bc)}{4c^2} \geq 0$$

it is also true.

\textbf{Case 2: $a^2+b^2+c^2 \geq 2(ab+bc+ca)$}, 

W.L.O.G assume $\textbf{c=max\{a,b,c\}}$, according to Cauchy-Schwarz inequality, $$\sqrt{a^2+bc} + \sqrt{b^2+ac} \leq \sqrt{2(a^2+b^2+ac+bc)}$$

and $$\sqrt{c^2+ab} \leq \sqrt{c^2+c^2} = \sqrt{2}c$$
$$LHS \leq \sqrt{2(a^2+b^2+ac+bc)} + \sqrt{2}c \leq {3\over2}(a+b+c)$$
$$\Longleftrightarrow \sqrt{2(a^2+b^2+ac+bc)} \leq {3\over2}(a+b+c) - \sqrt{2}c$$

$$\frac{a(b+c)}{2} \le \frac{(b+c)^2}{4} +\frac{a^2}{4} + \frac{7bc}{2} - \frac{3bc(b+c)}{2a} + \frac{(bc)^2}{4a^2}
$$

\section{Random forest}
\begin{CJK}{UTF8}{gbsn}
随机森林有n棵树，每棵树的准确率为60\%，请问最后随机森林的准确率为多少。
\end{CJK}
\newline\textbf{Solution:}
\\Assume rule: few obey majority, we have:
$$P = \sum_{i=1+n/2}^{n}C_n^i0.6^i(1-0.6)^{n-i}$$

\section{Finite products of sine function}
\subsection{1}
Prove that $$\prod_{k=1}^{n-1}sin\frac{k\pi}{n} = \frac{n}{2^{n-1}} $$
\\\textbf{Proof.}

Consider $z^n=1$, each root is 
$$\xi_k = cos\frac{2k\pi}{n} + isin\frac{2k\pi}{n} = e^{i\frac{2k\pi}{n}}, k=0,1,2,...,n-1 $$
So we have
$$ z^n -1 = \prod_{k=0}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-\xi_0)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-1)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow z^{n-1}+...+z^2+z+1 = \prod_{k=1}^{n-1}(z-\xi_k)$$
By substituting  z=1 ,$\Longrightarrow n = \prod_{k=1}^{n-1}(1-\xi_k) $

Next, modulus for both sides,
$$ |n| = n = |\prod_{k=1}^{n-1}(1-\xi_k)| = \prod_{k=1}^{n-1}|(1-\xi_k)|$$
$$ 1 - \xi_k = 1-(cos\frac{2k\pi}{n} + isin\frac{2k\pi}{n}) = 2sin\frac{k\pi}{n}(sin\frac{k\pi}{n} -icos\frac{k\pi}{n})$$
$$ |1 - \xi_k| = 2sin\frac{k\pi}{n} $$
So, 
$$ n = 2^{n-1}\prod_{k=1}^{n-1}sin\frac{k\pi}{n}$$
$$\prod_{k=1}^{n-1}sin\frac{k\pi}{n} = \frac{n}{2^{n-1}} $$

\subsection{2}
\textbf{Prove } $sin\frac{\pi}{4n}\cdot sin\frac{3\pi}{4n}\cdot sin\frac{5\pi}{4n}\cdots sin\frac{(2n-1)\pi}{4n} = (\frac{1}{\sqrt{2}})^{2n-1}$ 
\\\textbf{Proof.}

Let 
$$P=sin\frac{\pi}{4n}\cdot sin\frac{3\pi}{4n}\cdot sin\frac{5\pi}{4n}\cdots sin\frac{(2n-1)\pi}{4n}$$
$$Q=sin\frac{(2n+1)\pi}{4n}\cdot sin\frac{(2n+3)\pi}{4n}\cdots sin\frac{(4n-1)\pi}{4n}$$
Obviously, P=Q, so
$$ P^2 = PQ = \prod_{k=0}^{2n-1}sin\frac{(2k+1)\pi}{4n}$$
Consider $Z^{2n} +1 = 0$, each root was 
$$ \xi_k = cos\frac{(2k+1)\pi}{2n} + isin\frac{(2k+1)\pi}{2n}= e^{i\frac{(2k+1)\pi}{2n}}$$
So, we have
$$ Z^{2n} + 1 = \prod_{k=0}^{2n-1}(Z-\xi_k) $$
By substituting z=1, and take modulus on both sides, 
$$ 2 = \prod_{k=0}^{2n-1}|1-\xi_k| = \prod_{k=0}^{2n-1}2sin\frac{(2k+1)\pi}{4n} = 2^{2n}\cdot P^2$$
$$P=(\frac{1}{\sqrt{2}})^{2n-1}$$

\section{Calculus}
\subsection{Derivative}
\begin{enumerate}

	\item Find the first derivative of $y=x^x$, and thus $y=f(x)^{g(x)}$
	\\\textbf{Solution.}
	\\\textbf{Note} that the function defined by $y = x^x$ is neither a power function of the form $x^n$ nor an exponential function of the form $a^x$ and the formulas of Differentiation of these functions cannot be used. We need to find another method to find the first derivative of the above function. 
	
	\begin{align*}
		y  &= x^x        \\
		\ln y &= x \ln x \\
	\frac{y'}{y} &= 1+\ln{x} \\
	y' &= y(1+\ln{x}) = (1+\ln{x})x^x
	\end{align*}
	Same process for  $y=f(x)^{g(x)}$,
	\begin{align*}
		y &= f(x)^{g(x)} \\
		\ln{y} &= g(x) \ln{f(x)} \\
		\frac{y'}{y} &= g'(x) \ln{f(x)} + \frac{g(x)}{f(x)} f'(x) \\
		y' &= y \cdot (g'(x) \ln{f(x)} + \frac{g(x)}{f(x)} f'(x) ) = f(x)^{g(x)} \cdot \left( g'(x) \ln{f(x)} + \frac{g(x)}{f(x)} f'(x) \right)
	\end{align*}
	\item good morning....
	
\end{enumerate}

\section{Real analysis}
\subsection{1}
Prove that the set of all rational numbers is countable.
\\\textbf{Proof.}
\\\textbf{First.} $Q^{+}$ is countable, that is $Q^{+} \sim N.$
\\Consider subset of N: 
$$ A=\{ 2^p3^q| p,q\in N\} $$

Define mapping  $Q^+ \longrightarrow A : $, rational number of $\frac{p}{q}$ map to $ 2^p3^q$ 

It is clearly a one-to-one, but $ A \subset N \subset Q+$, so :
$$ A \sim N \sim Q^+$$
$Q^+$ is countable.
\\\textbf{Second.} 
Obviously, $Q^- \sim Q^+ $ for map of $\frac{p}{q} \rightarrow -\frac{p}{q}$

Then 
$$ Q = Q^- \cup \{0\} \cup Q^+$$ is countable.

\section{Inequality}
\subsection{1}
For nonnegative real number $x \ge 0$, prove that
$$ \sum_{k=1}^{n}\sqrt{x^2+2xcos\frac{(2k-1)\pi}{2n} + 1} \ge 1+ \sum_{k=1}^{n-1}\sqrt{x^2+2xcos\frac{2k\pi}{2n} + 1}$$
where equality holds when $x=0$.
\\\textbf{Proof.}

Obviously, inequality hold for $x \ge 1$, in the following, assume $0 \le x \le 1$,
$$RHS=1+ \sum_{k=1}^{n}\sqrt{x^2+2xcos\frac{2k\pi}{2n} + 1} - \sqrt{x^2-2x+1}$$
$$=x+\sum_{k=1}^{n}\sqrt{x^2+2xcos\frac{2k\pi}{2n} + 1}$$

which transform the problem into proving ( for $ x \le 1$):
$$\sum_{k=1}^{n}\sqrt{x^2+2xcos\frac{(2k-1)\pi}{2n} + 1} - \sum_{k=1}^{n}\sqrt{x^2+2xcos\frac{2k\pi}{2n} + 1} \ge x$$
$$\Longleftrightarrow \sum_{k=1}^{n}\frac{2x(cos\frac{(2k-1)\pi}{2n})-cos\frac{2k\pi}{2n}}{\sqrt{x^2+2xcos\frac{(2k-1)\pi}{2n} + 1} + \sqrt{x^2+2xcos\frac{2k\pi}{2n} + 1}} \ge x$$
Let $$H(x) = \sum_{k=1}^{n}\frac{(cos\frac{(2k-1)\pi}{2n})-cos\frac{2k\pi}{2n}}{\sqrt{x^2+2xcos\frac{(2k-1)\pi}{2n} + 1} + \sqrt{x^2+2xcos\frac{2k\pi}{2n} + 1}}$$,
\\H(x) is a monotonically decreasing function on interval [$cos\frac{\pi}{2n}$,1], so $H(x) \ge H(1)$,

$$H(1) = \sum_{k=1}^{n}[cos\frac{(2k-1)\pi}{4n} - cos\frac{2k\pi}{4n}] $$
$$=\sum_{k=1}^{n}[cos\frac{(2k-1)\pi}{4n} - cos\frac{2k\pi}{4n}]$$
$$\Longleftrightarrow 2\sum_{k=1}^{n}[sin^2\frac{(2k)\pi}{8n} - sin^2\frac{(2k-1)\pi}{8n}] \ge \frac{1}{2}$$

$$ \sum_{k=1}^{n/2}\sqrt{x^2-2xcos\frac{(2k-1)\pi}{n} + 1} +x \ge \sum_{k=1}^{n/2}\sqrt{x^2-2xcos\frac{2k\pi}{n} + 1}$$
where equality holds when $x=0$.


\section{Number Theory}
\subsection{1}
First, for $0 \le k \le 12$, only $2^{12} \equiv 2^0 \equiv 1 \quad mod(13)$ 



\begin{tabular}{c|c}
	\hline
	\hline i & $2^i$ mod 13\\
	\hline
	 1&2 \\
	 2&4 \\
	 3&8 \\
	 4&3 \\
	 5&6 \\
	 6&12 \\
	 7&11 \\
	 8&9 \\
	 9&5 \\
	 10&10 \\
	 11&7 \\
	 12&1 \\
	
	\hline
\end{tabular}

Assume $k=12i+j$, $0 \le j < 12$, then

$$2^k \equiv 2^{12i+j} \equiv 2^{12i} \cdot 2^j \equiv 1\cdot 2^j \equiv 1 \quad mod(13)$$

So $2^j=1$, $j=0$, and $k=12i, k|12$

\section{Fun Mathematica}
\subsection{1}
Find value of : $$\sqrt[3]{7-\sqrt{50}} + \sqrt[3]{7+\sqrt{50}}$$

$$\sqrt[\leftroot{-3}\uproot{-3}3]{\dfrac{333}{444}}$$

\section{Evaluation}
\begin{CJK}{UTF8}{gbsn}
	content...
这里是中文内容, 哈哈! Good day!
你好\citep{abley2006operation}再见。

This is a method! good\citep{wang2010research} day!

正方形 $\mathit{ABCD}$，角 $\mathit{ABC}$，$\mathit{EF} = \mathit{EG}$

\end{CJK}

\section{high school}
\subsection{inequality}
For positive number $x,y,z >0$, such that 
\[ x+y+z+2xyz = 1+xy+yz+zx\]

\textbf{Prove } that ,\[x^3+y^3+z^3+5xyz \ge 1\]

\textbf{Proof.}
$$	x^3+y^3+z^3+5xyz \ge 1 \Longleftrightarrow (x+y+z)[(x+y+z)^2-3(xy+yz+zx)] + 8xyz \ge 1 \quad \textbf{(*)}$$
thus, let $$u=x+y+z, v=xyz$$
$$\Longleftrightarrow u[u^2-3(u+2v-1)] + 8v \ge 1$$
$$\Longleftrightarrow u^3-3vu + 8v-1 \ge 0 $$
$$\Longleftrightarrow u^3-3u^2-3(2v-1)u + 8v-1 \ge 0 \quad \textbf{(**)}$$

But, \[ x+y+z+2xyz = 1+xy+yz+zx, \Longleftrightarrow (1-x)(1-y)(1-z) = xyz\]

thus let $x=\frac{a}{a+b},y=\frac{b}{b+c},z=\frac{c}{c+a}$,

which gives $u=x+y+z\in(1,2), v \in(0,\frac{1}{8})$

and (**) is obvious.

\subsection{problem}
$$1+\frac{1}{1+1}+\frac{1}{1+2}+...+\frac{1}{1+2+...+2000}$$
$$=\frac{1}{1+1}+2(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2000\times2001})=\frac{10001}{4002}$$
\paragraph{hh}
$$\frac{1}{1\times2}+\frac{2}{1\times2\times3}+...\frac{6}{1\times2\times3\times4\times5\times6\times7}$$
$$=1-\frac{1}{1\times2} + \frac{1}{1\times2}-\frac{1}{1\times2\times3}+...=1-\frac{1}{1\times2\times3\times4\times5\times6\times7}=\frac{5039}{5040}$$

$$\frac{sin(x-43^{\circ})}{sinx\cdot sin73^{\circ}}=2$$
$$sin(x-43^{\circ}) = 2sinx\cdot sin(30^{\circ}+43^{\circ})$$
$$sinx\cdot cos43^{\circ}-cosx\cdot sin43^{\circ}=2sinx(\frac{cos43^{\circ}}{2}+\frac{\sqrt{3}sin43^{\circ}}{2})$$
$$tanx=-\frac{1}{\sqrt{3}}$$
Which gives $x=150^{\circ}$

$$1=tan15^{\circ}\cdot tan21^{\circ}\cdot tan27^{\circ}\cdot tan87^{\circ} $$
$$\Longleftrightarrow sin15^{\circ}\cdot sin21^{\circ}\cdot sin27^{\circ}\cdot sin87^{\circ} = cos15^{\circ}\cdot  cos21^{\circ}\cdot cos27^{\circ}\cdot cos87^{\circ}$$
$$\Longleftrightarrow [cos(15^{\circ}-21^{\circ})-cos(15^{\circ}+21^{\circ})][cos(27^{\circ}-87^{\circ})-cos(27^{\circ}+87^{\circ})]$$
$$=[cos(15^{\circ}-21^{\circ})+cos(15^{\circ}+21^{\circ})][cos(27^{\circ}-97^{\circ})+cos(27^{\circ}+87^{\circ})]$$
$$\Longleftrightarrow cos6^{\circ}\cdot cos114^{\circ} + cos60^{\circ}\cdot cos36^{\circ}=0$$
$$\Longleftrightarrow \frac{1}{2}(cos120^{\circ}+cos108^{\circ}+cos96^{\circ}+cos24^{\circ})=0$$
$$\Longleftrightarrow cos120^{\circ}-sin18^{\circ}+2cos60^{\circ}\cdot cos36^{\circ}=0 $$
$$\Longleftrightarrow cos36^{\circ} - sin18^{\circ} -\frac{1}{2}=0 \quad\quad  (*)$$

But,
$$\sin18^{\circ}=\frac{\sqrt{5}-1}{4},\quad cos36^{\circ}=1-2sin^218^{\circ}=\frac{\sqrt{5}+1}{4}$$
So, (*) is clearly true.

\begin{equation}
x^2 + y^2 = 1
\end{equation}

\[ sinx, \sin x\cos x \tan x, 36\equiv5\mod{31}, 25\equiv4\pmod{22} \]

\subsection{Natural number breakup}

\begin{CJK}{UTF8}{gbsn}
	
\begin{eqnarray*}
\textbf{3465} & =1732+1733 \quad \text{(2项)} \\
     & =1154+1155+1156 \quad \text{(3项)} \\
     & =691 +......... \quad \text{(5项)} \\
     & =575 +.........  \quad \text{(6项)} \\
     & =492 +.........  \quad \text{(7项)} \\
     & =381 +.........  \quad \text{(9项)} \\
     & =342 +.........  \quad \text{(10项)} \\
     & =310 +.........  \quad \text{(11项)} \\     
\\
\textbf{180180} & = 4 \times 5 \times 7 \times 9 \times 11 \times 13 \\
	& = 3,5,7,8,9,11,13,15 \text{项}
\end{eqnarray*}

\end{CJK}
\newpage
\section{Algorithm}
\subsection{Interview}
\begin{enumerate}
	\item You are given two eggs, and access to a 100-storey building. Both eggs are identical. The aim is to find out the highest floor from which an egg will not break when dropped out of a window from that floor. If an egg is dropped and does not break, it is undamaged and can be dropped again. However, once an egg is broken, that’s it for that egg. 
	

	If an egg breaks when dropped from floor n, then it would also have broken from any floor above that. If an egg survives a fall, then it will survive any fall shorter than that.

	
	The question is: What strategy should you adopt to minimize the number egg drops it takes to find the solution?. (And what is the worst case for the number of drops it will take?)


	There are no tricks, gotchas or other devious ruses. Don’t rat-hole with issues related to terminal velocity, potential energy or wind resistance. This is a math puzzle plain and simple.
	Think about the puzzle for a few minutes, and then read on. 
\\\textbf{Solution.} \\

\end{enumerate}
\newpage
\section{2017-BEIKING Univ. and TSinghua Univ.}
\subsection{PKU}
\begin{CJK}{UTF8}{gbsn}
	
\textbf{2017北京大学数学科学学院秋令营10月13日}
\\
3.给定素数p, 正整数n,a, 其中(a,p)=1, 证明: 存在无穷多个正整数k,使得 $p^n|k^k-a$.\\
\textbf{Proof.}\\
(1) if a=1, let $k=t\cdot p^n + 1,t=1,2,3,...$\\
Thus, $k^k\equiv 1 \pmod{p^n}$,   \\
Which gives $p^n|k^k-1=k^k-a$  \\
(2) assume $a>1$, \\
According to Euler's Theorem, we have: $a^s \equiv 1 \pmod{p^n}$ , where $s=\phi{(p^n)}$\\
so, $$a^s-1 = r_1\cdot q^n$$
$$ (a-1)(a^{s-1}+...+a+1) = r_1\cdot q^n$$
$$a-1=r\cdot q^n$$
\\
let $k=t\cdot (a-1) + 1, t=1,2,3,\cdots $\\
Thus, $k^k \equiv 1 \pmod{a-1}$ \\
$$ $$
\newpage
\subsection{THU}
5. 给定奇素数$p$, 求集合$\left\{(x,y)\mid x^2+y^2 \equiv a \pmod{p}; x,y\in \{0,1,2,\cdots ,(p-1)\} \right\}$ 的元素个数。
\\
\textbf{Solution.} \\
首先, $a\ne 0$时, 方程$uv\equiv a \pmod p $的解$(u,v)$的数目是 $p-1$。\\
原方程可以写为: $x^2 \equiv a-y^2 \pmod{p}$, 当$y$ 依次取遍 $\{0,1,2,...,p-1\}$时, 考虑$x$所有的可能取值。\\
记 某个元$b$模$p$的逆元$c=b^{-1}$, 也即$bc\equiv b\cdot b^{-1}\equiv1\pmod{p}$\\
\\
(1) 当$p\equiv1 \pmod{4}$时，由欧拉判别法，若$b$是模$p$的平方剩余, 则$-b$也是模$p$的平方剩余, 于是平方剩余是以$\pm b$的形式成对出现。\\
从而$-1$是模p的平方剩余，可设$z^2\equiv -1 \pmod{p}$, 

(i) 如果$a=0$, 那么$x^2\equiv -y^2 \pmod{p}$, 由于平方剩余以$\pm b$形式成对出现,因此$x=y=0$，或者$x,y\ne 0$以$2(p-1)$组出现,
此时解的总数为: $1+2(p-1)=2p-1$

(ii) 如果$a\ne 0$, 那么有:
\begin{eqnarray*}
&x^2+y^2 \equiv a \pmod{p} & \Longrightarrow x^2 - (-1)y^2 \equiv a \pmod{p}\\
&\Longrightarrow x^2 - (zy)^2 \equiv a \pmod{p} & \Longrightarrow (x-zy)(x+zy) \equiv a \pmod{p}\\
&\Longrightarrow u\cdot v \equiv a \pmod{p}
\end{eqnarray*}
其中,$(u,v)$ 和 $(x,y)$一一对应.
\begin{eqnarray*}
	u=x-zy,v=x+zy \\
	\Longleftrightarrow x=(u+v)2^{-1}, y=(u-v)2^{-1}z^{-1}
\end{eqnarray*}
不同的$(u,v)$一共有$p-1$组, 所以不同的解$(x,y)$的数目也是 $p-1$.
\\
(2) 当$p\equiv3 \pmod{4}$时，由欧拉判别法, 如果$b$是平方剩余, 那么$-b$是平方非剩余,

(i) 如果$a=0$, 那么$x^2+y^2 \equiv 0 \pmod p$，只有一组解: $(x,y)=(0,0)$

(ii) 如果$a\ne 0$, 当$y$ 依次取遍 $\{0,1,2,...,p-1\}$时, 考虑$x$的可能取值时, 对每个$y$,下述方程只有一个有解:\\
\begin{eqnarray*}
	x^2 \equiv a - y^2 \pmod p \\
	x^2 \equiv y^2 - a \pmod p \\
\end{eqnarray*}
因为所有的$y$的数目是$p$, 所以上述两个方程给出的所有解$(x,y)$的数目是$2p$, 但是方程:\\
$$x^2 \equiv y^2 - a \pmod p \Longleftrightarrow x^2 - y^2 \equiv - a \equiv b\pmod p$$
解的数目是 $p-1$组， 故 $x^2 \equiv a - y^2 \pmod p$，也即$x^2 +  y^2\equiv a  \pmod p$ 的解的数目是: $2p-(p-1) = p+1$ 。\\
注： 如果$y^2 \equiv a \pmod p$，那么两个方程同时给出解$x=0$, 此时不同的$y$只有2个,所以总的解数目还是$2p$。\\
\\
综上, 解数目:
$$ =\left\{
\begin{aligned}
2p-1, & \quad a=0, p \equiv 1 \pmod 4 \\
1, &\quad  a=0, p \equiv 3 \pmod 4 \\
p-1, &\quad  a\ne 0, p \equiv 1 \pmod 4 \\
p+1, &\quad  a\ne 0, p \equiv 3 \pmod 4 \\
\end{aligned}
\right.
$$

\end{CJK}
\newpage
\section{Conclusion}
\begin{table}
	\begin{tabular}{l|l|c}
		\hline
		\hline 
		Host & Rib File & \# IPv4 entries\\
		\hline
		route-views2&rib.20160320.2200.bz2&630744\\
		sydney&rib.20160320.2000.bz2&602715\\
		saopaulo&rib.20160329.1600.bz2&613371\\
		wide&rib.20160630.2000.bz2&601203\\
		linx&rib.20160628.0200.bz2&618519\\
		
		\hline
	\end{tabular}
	\caption{real routing table from Route Views project}
	\label{tab:rvdata}
\end{table}

I always thought something was fundamentally wrong with the universe \citep{adams1995hitchhiker}
\begin{figure}[h!]
	\centering
	\includegraphics[scale=1.7]{universe.jpg}
	\caption{The Universe}
	\label{fig:univerise}
\end{figure}

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